A . 3x + 2(x + 1) = 6x - 7

<=> 3x + 2x + 2 = 6x -7

<=> 5x - 6x = -7 - 2

<=> -x = -9

<=> x =9

B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)

=> 3(x +3) < 5(5 -x)

<=> 3x+9 < 25 - 5x

<=> 3x + 5x < 25 - 9

<=> 8x < 16

<=> x < 2

C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)

<=> 5(x - 4) + 2x = 2(x +1)

<=> 5x - 20 + 2x = 2x + 2

<=>7x - 2x = 2 + 20

<=> 5x = 22

<=> x =\(\frac{22}{5}\)

\(\frac{3x}{5}+\frac{x-1}{4}=5-\frac{3x-1}{2}\)

\(\Leftrightarrow\frac{3x.4+5\left(x-1\right)}{20}=\frac{20.5-10\left(3x-1\right)}{20}\)

\(\Rightarrow12x+5x-5=100-30x+10\)

\(\Leftrightarrow47x=115\)

\(\Leftrightarrow x=\frac{115}{47}\)

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

a: ĐKXĐ: 2-|x-2|>=0

=>|x-2|<=2

=>-2<=x-2<=2

=>0<=x<=4

TH1: 0<=x<2

=>x-2<0

mà \(\sqrt{2-\left|x-2\right|}>=0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{2-\left|x-2\right|}\) >x-2 với mọi x thỏa mãn

=>NHận

=>0<=x<2

TH2: 2<=x<=4

\(\sqrt{2-\left|x-2\right|}>x-2\)

=>\(2-\left|x-2\right|\ge\left(x-2\right)^2\)

=>\(\left(x-2\right)^2\le2-\left(x-2\right)=2-x+2=4-x\)

=>\(x^2-4x+4\le4-x\)

=>\(x^2-3x\le0\)

=>x(x-3)<=0

=>0<=x<=3

=>2<=x<=3

Vậy: 0<=x<=3

b: \(x^2+3x+2\ge2\cdot\sqrt{x^2+3x+5}\)

=>\(x^2+3x+5-2\sqrt{x^2+3x+5}-3\ge0\)

=>\(\left(\sqrt{x^2+3x+5}-3\right)\left(\sqrt{x^2+3x+5}+1\right)>=0\)

=>\(\sqrt{x^2+3x+5}-3\ge0\)

=>\(\sqrt{x^2+3x+5}\ge3\)

=>\(x^2+3x+5\ge9\)

=>\(x^2+3x-4\ge0\)

=>(x+4)(x-1)>=0

=>x>=1 hoặc x<=-4

\(\frac{{3x - 2}}{5} + \frac{3}{2} = \frac{{4 - x}}{{10}}\)

\(\frac{{\left( {3x - 2} \right).2}}{{5.2}} + \frac{{3.5}}{{2.5}} = \frac{{4 - x}}{{10}}\)

\(\frac{{6x - 4}}{{10}} + \frac{{15}}{{10}} = \frac{{4 - x}}{{10}}\)

\(6x - 4 + 15 = 4 - x\)    

\(6x + x = 4 + 4 - 15\)

\(7x =  - 15\)

\(x = \left( { - 15} \right):7\)

\(x = \frac{{ - 15}}{7}\)

Vậy phương trình có nghiệm là \(x = \frac{{ - 15}}{7}\).

a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x²
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :

\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)

\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)

Đến đây ta đặt  \(x+\frac{60}{x}+16=t\left(1\right)\)

Ta được :

\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)

Từ đó ta lắp vào ( 1 ) tính được x 

\(\frac{2x-1}{3x^2+7x+2}+\frac{3}{9x^2+15x+4}-\frac{2x+7}{3x^2-5x-12}=\frac{5}{x+2}\)

\(\Leftrightarrow\frac{2x-1}{\left(3x+1\right)\left(x+2\right)}+\frac{3}{\left(3x+1\right)\left(3x+4\right)}-\frac{2x+7}{\left(4x+3\right)\left(x-3\right)}=\frac{5}{\left(x+2\right)}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{3x+1}+\frac{1}{3x+1}-\frac{1}{3x+4}+\frac{1}{3x+4}-\frac{1}{x-3}=\frac{5}{x+2}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x-3}=\frac{5}{x+2}\)

\(\Leftrightarrow\frac{x-3-x-2}{\left(x+2\right)\left(x-3\right)}=\frac{5\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow5x-3=-5\)

\(\Leftrightarrow x=-\frac{2}{5}\)

Chúc bạn học tốt !!!